Tree Diagrams in Probability Calculations | 树形图在概率计算中的应用方法

📚 Tree Diagrams in Probability Calculations | 树形图在概率计算中的应用方法

Tree diagrams are a powerful visual tool used to calculate probabilities in multi-stage random experiments. They help us organise possible outcomes and their associated probabilities in a clear, structured way, making complex probability problems much more manageable.

树形图是概率计算中一种强大的可视化工具，特别适用于多阶段随机试验。它能帮助我们以清晰、结构化的方式整理所有可能的结果及其对应概率，使复杂概率问题变得更加易于处理。

1. Introduction to Tree Diagrams | 树形图简介

A tree diagram consists of branches that represent each possible outcome at every stage of an experiment. Starting from a single point on the left, each branch extends to the right, splitting into further branches at each subsequent stage. Along each branch, we write the probability of that outcome occurring.

树形图由若干“树枝”组成，每条树枝代表试验中每个阶段可能出现的一种结果。从左侧的起点出发，每条树枝向右延伸，在后续每个阶段继续分叉。在每条树枝上，我们需要标明该结果发生的概率。

The key advantage of using tree diagrams is that they provide a systematic method to list all possible outcomes of an experiment. Instead of guessing or memorising formulas, you can simply “read” the probabilities from the diagram and multiply along the branches to find the probability of any combined outcome.

使用树形图的最大优势在于，它提供了一种系统化列出试验所有可能结果的方法。你不需要猜测或死记公式，只需从图中“读取”相关概率，沿着树枝相乘，即可求出任意复合结果发生的概率。

2. The Multiplication Rule | 乘法法则

When calculating probabilities using a tree diagram, the fundamental operation is multiplication. To find the probability of a particular sequence of outcomes (a path through the tree), we multiply the probabilities written on each branch along that path.

在使用树形图计算概率时，最基本的运算是乘法。要求某一特定结果序列（即从树根到终点的某一条路径）的概率，我们需要将这条路径上每条树枝所标注的概率相乘。

P(path) = P(first outcome) × P(second outcome) × …

For example, if we toss a fair coin twice, the probability of getting two heads is the product of the probability of heads on the first toss and the probability of heads on the second toss: ½ × ½ = ¼.

例如，抛一枚公平硬币两次，两次都为正面的概率等于第一次为正面的概率乘以第二次为正面的概率：½ × ½ = ¼。

This multiplication rule is the heart of every tree diagram calculation. It works because each branch probability represents a conditional probability given what has happened earlier in the tree.

这个乘法法则是所有树形图计算的核心。它之所以成立，是因为每条树枝上的概率本质上是一个条件概率，它基于该树枝之前已经发生的事件。

3. Constructing a Tree Diagram | 构建树形图的步骤

To construct a tree diagram accurately, follow these systematic steps:

要准确构建树形图，请遵循以下系统步骤：

Identify all stages of the experiment. Each stage will correspond to a column of branches in the tree.
确定试验的所有阶段。每个阶段对应树形图中一列树枝。
At each stage, list every possible outcome and write the corresponding probability on each branch.
在每个阶段，列出所有可能的结果，并在每根树枝上写出相应的概率。
For dependent events, ensure the probabilities on the second-stage branches reflect the condition of the first stage.
对于非独立事件，请确保第二阶段树枝上的概率反映第一阶段的结果条件。
At the end of each complete path, write the combined event and its probability (the product of all branch probabilities along the path).
在每条完整路径的末端，写出该路径对应的事件及其概率（即路径上所有树枝概率的乘积）。

Always check that the probabilities on all branches leaving any single point sum to 1. Every probability value must be between 0 and 1 inclusive.

请务必检查从任意一个节点出发的所有分支概率之和是否为 1。每个概率值都必须介于 0 和 1 之间（含端点）。

4. Independent Events | 独立事件

When two or more events are independent, the occurrence of one event does not affect the probability of the other. In a tree diagram, the branch probabilities at each stage remain the same regardless of which path was taken earlier.

当两个或多个事件相互独立时，一个事件的发生不会影响另一事件发生的概率。在树形图中，无论之前经过哪条路径，每个阶段的树枝概率都保持不变。

Example: Tossing a fair coin three times. At each toss, the probability of heads (H) is ½ and the probability of tails (T) is ½, no matter what outcomes occurred in previous tosses.

示例：将一枚公平硬币抛掷三次。每一次抛掷，正面（H）的概率都是 ½，反面（T）的概率也是 ½，不受之前抛掷结果的影响。

For independent events, the probability of a specific path is simply the product of the individual probabilities. For instance, the probability of H then T then H is ½ × ½ × ½ = ⅛.

对于独立事件，某条特定路径的概率就是各独立概率的乘积。例如，正、反、正这一序列的概率为 ½ × ½ × ½ = ⅛。

P(H then T then H) = ½ × ½ × ½ = ⅛

This simplicity makes tree diagrams especially useful for repeated independent trials, such as multiple coin tosses, dice rolls, or the same test repeated several times.

这种简洁性使得树形图特别适合处理重复的独立试验，例如多次抛硬币、掷骰子，或者同一测试重复进行多次的情况。

5. Dependent Events | 非独立事件（条件概率）

When events are dependent, the probability of a later event changes depending on what happened earlier. In a tree diagram, this is shown by writing different probabilities on branches that leave from different nodes at the same stage.

当事件相互依赖时，后续事件发生的概率会随着先前事件的结果而改变。在树形图中，这种依赖关系体现在同一阶段从不同节点出发的树枝上标注不同的概率。

Example: A bag contains 5 red balls and 3 blue balls. We draw two balls without replacement. The probability of drawing a blue ball on the second draw depends on whether the first ball was red or blue.

示例：一个袋子中装有 5 个红球和 3 个蓝球。我们从袋中不放回地抽取两个球。第二次抽到蓝球的概率取决于第一次抽到的是红球还是蓝球。

If the first ball is red, then 4 red and 3 blue remain, so the probability of blue on the second draw is 3/7. If the first ball is blue, then 5 red and 2 blue remain, so the probability of blue on the second draw is 2/7.

如果第一次抽到红球，袋中剩余 4 红 3 蓝，因此第二次抽到蓝球的概率为 3/7。如果第一次抽到蓝球，袋中剩余 5 红 2 蓝，因此第二次抽到蓝球的概率为 2/7。

These conditional probabilities must be written on the appropriate second-stage branches. Failing to adjust them is one of the most common errors students make.

这些条件概率必须标注在对应的第二阶段树枝上。忘记调整这些概率是学生最常犯的错误之一。

6. Worked Example: Selecting Balls | 例题解析：取球问题

Let us solve a complete problem using a tree diagram. A bag contains 3 red balls and 2 green balls. We draw two balls one after another without replacement. Find the probability that both balls are red.

让我们用树形图完整地解决一个问题。一个袋子中装有 3 个红球和 2 个绿球。我们不放回地先后抽取两个球。求两个球都是红色的概率。

First, we construct the tree:

首先，我们构建树形图：

First draw: P(R) = 3/5, P(G) = 2/5
第一次抽取：P(红) = 3/5，P(绿) = 2/5
Second draw given first is R: P(R) = 2/4, P(G) = 2/4
第一次为红时第二次抽取：P(红) = 2/4，P(绿) = 2/4
Second draw given first is G: P(R) = 3/4, P(G) = 1/4
第一次为绿时第二次抽取：P(红) = 3/4，P(绿) = 1/4

To find P(both red), we follow the path R then R and multiply the branch probabilities:

要求 P(两次都为红)，我们沿着红→红的路径，将各分支概率相乘：

P(R and R) = (3/5) × (2/4) = 6/20 = 3/10

Thus, the probability that both balls are red is 3/10. This example demonstrates how the tree diagram automatically handles the changing probabilities caused by sampling without replacement.

因此，两个球都是红色的概率为 3/10。这个例子展示了树形图如何自动处理由无放回抽样引起的概率变化。

7. Worked Example: Multiple Trials | 例题解析：多次试验

Consider a different situation: the probability that a student passes a driving test is 0.6 on any attempt. The student takes the test at most twice. Find the probability that the student passes on either the first or second attempt.

考虑另一种情况：一名学生任意一次驾照考试通过的概率为 0.6。该学生参加考试最多两次。求该学生在第一次或第二次考试中通过的概率。

We construct a tree with two stages. At the first attempt, the branches are Pass (P) with probability 0.6 and Fail (F) with probability 0.4. If the student passes, the process stops. If the student fails, they take the test again, with the same probabilities for the second attempt.

我们构建一个两阶段的树形图。在第一次考试中，分支为通过（P）概率 0.6 和未通过（F）概率 0.4。如果学生通过了，过程结束。如果未通过，则参加第二次考试，第二次考试的分支概率相同。

The probability of passing on the first attempt is 0.6. The probability of failing first and then passing is:

第一次就通过的概率为 0.6。第一次未通过然后第二次通过的概率为：

P(F then P) = 0.4 × 0.6 = 0.24

Since these two events are mutually exclusive (the student cannot pass on the first attempt AND fail first then pass), we add their probabilities:

由于这两个事件互斥（学生不可能同时“第一次通过”且“第一次未通过而后第二次通过”），我们将它们的概率相加：

P(pass in at most two attempts) = 0.6 + 0.24 = 0.84

This example illustrates that tree diagrams help us break a problem into parts: sometimes we multiply (for a single path) and sometimes we add (for multiple mutually exclusive paths).

这个例子说明，树形图帮助我们分解问题：有时我们需要相乘（针对单一路径），有时我们需要相加（针对多条互斥路径）。

8. Using Tree Diagrams to Find Probabilities of Combined Events | 利用树形图求复合事件的概率

In probability problems, we often need the probability of an event that can occur in more than one way. For instance, “at least one success” or “exactly one red ball”.

在概率问题中，我们经常需要求某个事件发生的概率，而该事件可能以多种方式发生。例如“至少一次成功”或“恰好一个红球”。

When an event can happen through several different paths in the tree, we first calculate the probability of each individual path, then add these probabilities together. The addition is justified because different paths are mutually exclusive — a single experiment can follow only one path.

当一个事件可以通过树形图中多条不同路径发生时，我们首先计算每条路径的概率，然后将这些概率相加。这种加法是合理的，因为不同路径是互斥的——一次试验只能沿着一条路径进行。

Example: Using the previous bag of 3 red and 2 green balls, find the probability of getting exactly one red ball when drawing two without replacement.

示例：沿用之前装有 3 红 2 绿的袋子，求不放回抽取两个球时恰好一个红球的概率。

There are two paths that give exactly one red: R then G, or G then R. We calculate each:

能够产生“恰好一个红球”的路径有两条：红→绿，或绿→红。我们分别计算：

P(R then G) = (3/5) × (2/4) = 6/20
P(G then R) = (2/5) × (3/4) = 6/20

P(exactly one red) = 6/20 + 6/20 = 12/20 = 3/5

Thus, the probability of exactly one red ball is 3/5. Always remember: multiply along branches for each path, then add across paths when the event can occur in multiple ways.

因此，恰好一个红球的概率为 3/5。请务必记住：沿每条路径的分支相乘，当事件可通过多种方式发生时再在各路径之间相加。

9. Symmetry and Checking Your Work | 对称性与检验答案

A useful property of many tree diagrams is symmetry. For example, when drawing two balls from a bag, the probability of R then G and the probability of G then R are often equal, especially when the number of balls of each colour is such that the multiplication yields the same product.

许多树形图具有一个有用的性质：对称性。例如，从一个袋子中抽取两个球时，红→绿和绿→红这两条路径的概率往往相等，尤其是当每种颜色球的数量使得乘积恰好相同时。

You can use symmetry as a quick check: if two paths are structurally identical but with the order of outcomes swapped, their probabilities should be equal. If they are not, you have likely made a calculation error.

你可以利用对称性进行快速检验：如果两条路径结构相同，只是结果的顺序互换，那么它们的概率应当相等。如果不相等，你很可能犯了计算错误。

Another useful check: the sum of the probabilities of all final branches (i.e., all complete paths) must equal 1. This verifies that you have correctly accounted for every possible outcome of the experiment.

另一个有用的检验：所有最终分支（即所有完整路径）的概率之和必须等于 1。这可以验证你是否已正确涵盖试验的所有可能结果。

10. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Even experienced students make mistakes when using tree diagrams. Here are the most common pitfalls and how to avoid them:

即使是经验丰富的学生也会在使用树形图时犯错。以下是最常见的陷阱及避免方法：

Mistake \| 错误	Correction \| 改正
Forgetting to adjust probabilities for sampling without replacement.	Always update the denominator (and numerator if necessary) after each draw without replacement.
忘记在无放回抽样中调整概率。	每次无放回抽取后，务必更新分母（必要时更新分子）。
Adding probabilities when they should be multiplied (or vice versa).	Multiply along a single path; add across different paths (mutually exclusive events).
在需要相乘的地方相加（或反之）。	单一路径内相乘；不同路径间相加（互斥事件）。
Writing probabilities as unreduced fractions or decimals incorrectly.	Convert all probabilities to the same form (fraction or decimal) and simplify where possible.
将概率写为未约分的分数或错误的十进制数。	将所有概率统一为相同形式（分数或小数），并尽可能化简。
Omitting a branch that represents a possible outcome.	Before calculating, list all possible outcomes explicitly and ensure each appears as a branch.
遗漏表示某种可能结果的分支。	在计算前，明确列出所有可能结果，并确保每个结果都有对应的分支。

By being mindful of these common errors, you can significantly improve your accuracy when working with tree diagrams.

只要留意这些常见错误，你就能在使用树形图时显著提高准确率。

11. Practice Problems | 练习题

To master tree diagrams, regular practice is essential. Try these problems on your own, then check your answers.

要掌握树形图，定期练习至关重要。请先独立思考以下问题，再核对答案。

Problem 1: A die is rolled twice. Draw a tree diagram to find the probability that both rolls show a number greater than 4.

问题 1：一枚骰子掷两次。请画出树形图，求两次都掷出大于 4 的数字的概率。

Solution: P(greater than 4) = 2/6 = 1/3 on each roll. Since rolls are independent, P(both greater than 4) = (1/3) × (1/3) = 1/9.

解答：每次掷出大于 4 的概率为 2/6 = 1/3。由于两次掷骰相互独立，P(两次都大于 4) = (1/3) × (1/3) = 1/9。

Problem 2: A box contains 4 white and 6 black marbles. Two marbles are drawn without replacement. Find the probability that the two marbles are the same colour.

问题 2：一个盒子中有 4 个白球和 6 个黑球。无放回地抽取两个球。求两个球颜色相同的概率。

Solution: P(WW) = (4/10) × (3/9) = 12/90 = 2/15. P(BB) = (6/10) × (5/9) = 30/90 = 1/3. Therefore P(same colour) = 2/15 + 1/3 = 2/15 + 5/15 = 7/15.

解答：P(白白) = (4/10) × (3/9) = 12/90 = 2/15。P(黑黑) = (6/10) × (5/9) = 30/90 = 1/3。因此 P(同色) = 2/15 + 1/3 = 2/15 + 5/15 = 7/15。

Problem 3: A tennis player wins 70% of her matches. If she plays two matches, find the probability that she wins at least one match, assuming each match is independent.

问题 3：一位网球选手赢得比赛的胜率为 70%。如果她打两场比赛，假设每场比赛独立，求她至少赢一场比赛的概率。

Solution: P(both losses) = 0.3 × 0.3 = 0.09. Therefore P(at least one win) = 1 − 0.09 = 0.91.

解答：P(两场均输) = 0.3 × 0.3 = 0.09。因此 P(至少赢一场) = 1 − 0.09 = 0.91。

12. Summary | 小结

Tree diagrams are an essential tool for solving probability problems involving multiple stages. By drawing branches for each outcome and writing probabilities along them, you can systematically calculate the probability of any combination of events.

树形图是解决多阶段概率问题不可或缺的工具。通过为每个结果画出树枝并在其上标注概率，你可以系统地计算任意事件组合发生的概率。

The key rules are straightforward: multiply along branches for a single path, and add across paths when an event can occur in multiple ways. Always check that branch probabilities leaving the same node sum to 1, and adjust probabilities correctly when sampling without replacement.

核心规则非常简单：单一路径内相乘；当一个事件可通过多种方式发生时，在路径间相加。始终检查从同一节点出发的分支概率之和是否为 1，并在无放回抽样时正确调整概率。

With practice, you will find that tree diagrams not only give correct answers but also build deep intuition about how probability works in multi-stage experiments. This will serve you well in both examinations and real-world problem-solving.

通过练习，你会发现树形图不仅能给出正确的答案，还能帮助你深入理解多阶段试验中概率的运作原理。这将在考试和解决实际问题中为你带来巨大的帮助。

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Tree Diagrams in Probability Calculations | 树形图在概率计算中的应用方法